What is the extreme f(x)=(1-x^2)^2 ?

The extreme f(x)=(1-x^2)^2 is Minimum(-1,0),Maximum(0,1),Minimum(1,0)

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extreme f(x)=(1-x^2)^2

extreme points

f (x) = (1 - x^{2})^{2}

Minimum (- 1, 0), Maximum (0, 1), Minimum (1, 0)

Solution steps

f^{'} (x) = - 4 x (1 - x^{2})

Find intervals:

Decreasing

: - \infty < x < - 1,

Increasing

: - 1 < x < 0,

Decreasing

: 0 < x < 1,

Increasing

: 1 < x < \infty

Plug

x = - 1

into

(1 - x^{2})^{2} : 0

Minimum (- 1, 0)

Plug

x = 0

into

(1 - x^{2})^{2} : 1

Maximum (0, 1)

Plug

x = 1

into

(1 - x^{2})^{2} : 0

Minimum (1, 0)

Minimum (- 1, 0), Maximum (0, 1), Minimum (1, 0)

f (x, y) = 2 x y + x^{2} y - y

f (x) = 5 x^{\frac{2}{3}} - 2 x^{\frac{5}{3}}

f (x) = ln (x^{2} + x + 1)

f (x) = x^{2} - 4 x y + y^{3} + 4 y

V, V = 13 π r 2 h

What is the extreme f(x)=(1-x^2)^2 ?
The extreme f(x)=(1-x^2)^2 is Minimum(-1,0),Maximum(0,1),Minimum(1,0)